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DATABASE APPLICATIONS LABORATORY

Subject Code: 10CSL57 I.A. Marks : 25 Hours/Week : 03 Exam Hours: 03 Total Hours : 42 Exam Marks: 50

1. Consider the following relations:

Student (snum: integer, sname: string, major: string, level: string, age: integer) Class (name: string, meets at: string, room: string, d: integer) Enrolled (snum: integer, cname: string) Faculty (fid: integer, fname: string, deptid: integer) The meaning of these relations is straightforward; for example, Enrolled has one record per student-class pair such that the student is enrolled in the class. Level is a two character code with 4 different values (example: Junior: JR etc) Write the following queries in SQL. No duplicates should be printed in any of the answers.

i. Find the names of all Juniors (level = JR) who are enrolled in a class taught by Prof. Harshith ii. Find the names of all classes that either meet in room R128 or have five or more Students

enrolled. iii. Find the names of all students who are enrolled in two classes that meet at the same time. iv. Find the names of faculty members who teach in every room in which some class is taught. v. Find the names of faculty members for whom the combined enrollment of the courses that they

teach is less than five.

2. The following relations keep track of airline flight information: Flights (no: integer, from: string, to: string, distance: integer, Departs: time, arrives: time, price: real) Aircraft (aid: integer, aname: string, cruisingrange: integer) Certified (eid: integer, aid: integer) Employees (eid: integer, ename: string, salary: integer)

Note that the Employees relation describes pilots and other kinds of employees as well; Every pilot is certified for some aircraft, and only pilots are certified to fly.

Write each of the following queries in SQL.

i. Find the names of aircraft such that all pilots certified to operate them have salaries more than Rs.80, 000.

ii. For each pilot who is certified for more than three aircrafts, find the eid and the maximum cruisingrange of the aircraft for which she or he is certified.

iii. Find the names of pilots whose salary is less than the price of the cheapest route from Bengaluru to Frankfurt.

iv. For all aircraft with cruisingrange over 1000 Kms, .find the name of the aircraft and the average salary of all pilots certified for this aircraft.

v. Find the names of pilots certified for some Boeing aircraft. vi. Find the aids of all aircraft that can be used on routes from Bengaluru to New Delhi.

3. Consider the following database of student enrollment in courses & books adopted for each course.

STUDENT (regno: string, name: string, major: string, bdate:date) COURSE (course #:int, cname:string, dept:string) ENROLL ( regno:string, course#:int, sem:int, marks:int) BOOK _ ADOPTION (course# :int, sem:int, book-ISBN:int) TEXT (book-ISBN:int, book-title:string, publisher:string, author:string)

i. Create the above tables by properly specifying the primary keys and the foreign keys. ii. Enter at least five tuples for each relation.

iii. Demonstrate how you add a new text book to the database and make this book be adopted by some department.

iv. Produce a list of text books (include Course #, Book-ISBN, Book-title) in the alphabetical order for courses offered by the ‘CS’ department that use more than two books.

v. List any department that has all its adopted books published by a specific publisher. vi. Generate suitable reports.

vii. Create suitable front end for querying and displaying the results.

4. The following tables are maintained by a book dealer. AUTHOR (author-id:int, name:string, city:string, country:string) PUBLISHER (publisher-id:int, name:string, city:string, country:string) CATALOG (book-id:int, title:string, author-id:int, publisher-id:int, category-id:int, year:int, price:int) CATEGORY (category-id:int, description:string) ORDER-DETAILS (order-no:int, book-id:int, quantity:int)

i. Create the above tables by properly specifying the primary keys and the foreign keys. ii. Enter at least five tuples for each relation.

iii. Give the details of the authors who have 2 or more books in the catalog and the price of the books is greater than the average price of the books in the catalog and the year of publication is after 2000.

iv. Find the author of the book which has maximum sales. v. Demonstrate how you increase the price of books published by a specific publisher by 10%.

vi. Generate suitable reports. vii. Create suitable front end for querying and displaying the results.

5. Consider the following database for a banking enterprise

BRANCH(branch-name:string, branch-city:string, assets:real) ACCOUNT(accno:int, branch-name:string, balance:real) DEPOSITOR(customer-name:string, accno:int) CUSTOMER(customer-name:string, customer-street:string, customer-city:string) LOAN(loan-number:int, branch-name:string, amount:real) BORROWER(customer-name:string, loan-number:int)

i. Create the above tables by properly specifying the primary keys and the foreign keys ii. Enter at least five tuples for each relation

iii. Find all the customers who have at least two accounts at the Main branch. iv. Find all the customers who have an account at all the branches located in a specific city. v. Demonstrate how you delete all account tuples at every branch located in a specific city.

vi. Generate suitable reports. vii. Create suitable front end for querying and displaying the results.

Instructions:

1. The exercises are to be solved in an RDBMS environment like Oracle or DB2. 2. Suitable tuples have to be entered so that queries are executed correctly. 3. Front end may be created using either VB or VAJ or any other similar tool. 4. The student need not create the front end in the examination. The results of the queries may be

displayed directly. 5. Relevant queries other than the ones listed along with the exercises may also be asked in the

examination. 6. Questions must be asked based on lots.

DATABASE APPLICATIONS LABORATORY MIT V CSE/ISE

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COLLEGE DATABASE

Create table student2(

snum int primary key,

sname varchar(10) not null,

major varchar(10) not null,

lvl varchar(2) not null,

age int not null);

Create table faculty2(

fid int primary key,

fname varchar(15) not null,

dept_id int not null);

Create table class2(

cname varchar(10) primary key,

meetsat varchar(10) not null,

room varchar(10) not null,

d int references faculty2(fid));

Create table enrolled2 (

snum int references student2(snum),

cname varchar(10) references class2(cname),

primary key(snum,cname));

Describe student;

Name Null? Type

--------------------- -------- ---------------

SNUM NOT NULL NUMBER (38)

SNAME NOT NULL VARCHAR2 (10)

MAJOR NOT NULL VARCHAR2 (10)

LVL NOT NULL VARCHAR2 (2)

AGE NOT NULL NUMBER (38)


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Describe faculty;

Name Null? Type

---------------------------------- -------- -------------

FID NOT NULL NUMBER(38)

FNAME NOT NULL VARCHAR2(15)

DEPT_ID NOT NULL NUMBER(38)

Describe class;

Name Null? Type

NAME NOT NULL VARCHAR (10)

MEETSAT NOT NULL VARCHAR (10)

ROOM NOT NULL VARCHAR (10)

FID NOT NULL NUMBER(38)

Describe enrolled;

Name Null? Type

----------------------------------------- -------- ------------

SNUM NOT NULL NUMBER (38)

CNAME NOT NULL VARCHAR2 (10)

VALUES FOR TABLES:

Insert into student2 values ('111','john','cs','jr','19');

Insert into student2 values ('222','tarun','cse','sr','22');

Insert into student2 values ('333','mac','civil','jr','21');

Insert into student2 values ('444','ethan','mech','jr','20');

Insert into student2 values ('555','sonia','eee','sr','22');


[Type text] Page 3

SELECT * from student2;

SNUM SNAME MAJOR LVL AGE

----- ---------- ---------- -- ----------

111 john cse jr 19

222 tarun cse sr 22

333 mac civil jr 21

444 ethan mech jr 20

555 sonia eee sr 22

Insert into faculty2 values ('101','asthana','1');

Insert into faculty2 values ('102','sham','2');

Insert into faculty2 values ('103','abhishek','3');

Insert into faculty2 values ('104','vinay','4');

Insert into faculty2 values ('105','shekar','5');

Insert into faculty2 values ('106','harshith','6');

SELECT * from faculty2;

FID FNAME DEPT_ID

---- --------------- ----------

101 asthana 1

102 sham 2

103 abhishek 3

104 vinay 4

105 shekar 5

106 harshith 6

Insert into class2 values ('se','11am','r128', 106);

Insert into class2 values ('os','11am','r127', 102);

Insert into class2 values ('cn','2pm','r126', 106);

Insert into class2 values ('flat','8am','r125', 102);

Insert into class2 values ('dbms','11am','r127', 106);

Insert into class2 values ('ss','12pm','r128', 106);

Insert into class2 values ('ptw','8am','r128', 102);

Insert into class2 values ('oomd','11am','r127', 106);


[Type text] Page 4

SELECT * from class2;

CNAME MEETSAT ROOM D

---------- ---------- ------------- -------

se 11am r128 106

os 11am r127 102

cn 2pm r126 106

flat 8am r125 102

dbms 11am r127 106

ss 12pm r128 106

oomd 11am r127 106

ptw 8am r128 102

Insert into enrolled2 values ('111','se');

Insert into enrolled2 values ('222,'dbms');

Insert into enrolled2 values ('111','dbms');




SELECT * from enrolled2;

SNUM CNAME

-------- ----------

111 se

333 dbms

444 dbms

555 dbms

111 dbms


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i)Find the names of all juniors (level = JR) who are enrolled in a class taught by

Prof. Harshith.

SELECT distinct sname

FROM student s,class c,enrolled e,faculty f

WHERE lvl='jr' and

s.snum=e.snum and

c.cname=e.cname and

fid=d and

fname='harshith';

SNAME

------

ethan

john

mac

ii. Find the names of all classes that either meet in room R128 or have five or more

Students enrolled.

SELECT cname

FROM class2

WHERE room='r128' or

cname in (SELECT cname

FROM enrolled2

GROUP BY cname

HAVING count (snum)>=5); CNAME

-----

se

ss

ptw


[Type text] Page 6

iii. Find the names of all students who are enrolled in two classes that meet at the

same time.

SELECT sname

FROM student2

WHERE snum in (SELECT snum

FROM enrolled2 e, class2 c

WHERE meetsat='11am' and

c.cname=e.cname

GROUP BY snum

HAVING count (snum) =2);

SNAME

---------

John

iv. Find the names of faculty members who teach in every room in which some

class is taught.

SELECT fname

FROM faculty2 f, class2 c,enrolled2 e

WHERE fid=d and

c.cname=e.cname

GROUP BY fname

HAVING count (distinct room) = (SELECT count (distinct room)

FROM class2 c,enrolled2 e

WHERE c.cname=e.cname);

Fname

-----

harshith

v. Find the names of faculty members for whom the combined enrollment of the

courses that they teach is less than five.

SELECT fname

FROM faculty2 f, class2 c

WHERE fid=d


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GROUP BY fname

HAVING count (cname)<5;

FNAME

--------

Sham


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AIRLINE DATABASE

create table aircraft2(

aid int primary key,

aname varchar(10),

cruisingrange int);

create table employees2(

eid int primary key,

ename varchar(10),

salary real);

create table certified2(

eid int,

aid int,

foreign key (eid) references employees2(eid),

foreign key (aid) references aircraft2(aid),

primary key(eid,aid));

create table flights2(

num int primary key ,

frome varchar(10),

too varchar(10),

distance int,

departs int,

arrives int,

price real);

Desc aircraft;

Name Null? Type

----------------------------- -------- -------------

AID NOT NULL NUMBER(38)

ANAME VARCHAR2(10)

CRUISINGRANGE NUMBER(38)


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Desc employees;

Name Null? Type

---------- -------- -------------

EID NOT NULL NUMBER(38)

ENAME VARCHAR2(10)

SALARY FLOAT(63)

Desc certified;

Name Null? Type

--------------- -------- -----------

EID NOT NULL NUMBER(38)

AID NOT NULL NUMBER(38)

Desc flights;

Name Null? Type

----------------- -------------

NUM NOT NULL NUMBER(38)

FROME VARCHAR2(10)

TOO VARCHAR2(10)

DISTANCE NUMBER(38)

DEPARTS NUMBER(38)

ARRIVES NUMBER(38)

PRICE FLOAT(63)

VALUES FOR TABLES:

Insert into aircraft2 values(50,'kingfisher',20000);

Insert into aircraft2 values(51,'jet',500);

Insert into aircraft2 values(52,'IA',25000);

Insert into aircraft2 values(53,'Air india',30000);

Insert into aircraft2 values(54,'Boeing3',26000);

Insert into aircraft2 values(55,'Boeing5',25000);


[Type text] Page 10

Select * from aircraft;

AID ANAME CRUISINGRANGE

---- ---------- -------------

50 kingfisher 20000

51 jet 500

52 IA 25000

53 Air india 30000

54 boeing3 26000

55 boeing5 25000

AID ANAME CRUISINGRANGE

50 kingfisher 20000

51 jet 500

52 IA 25000

53 Air india 30000

54 boeing3 26000

55 boeing5 25000

Insert into employees2 values(200,'manish',100000);

Insert into employees2 values(201,'ramesh',200000);

Insert into employees2 values(202,'ganesh',4000);

Insert into employees2 values(203,'suresh',85000);

Insert into employees2 values(204,'mahesh',80000);

Select * from employees;

EID ENAME SALARY

--- ---------- ----------

200 manish 100000

201 ramesh 200000

202 ganesh 4000

203 suresh 85000

204 mahesh 80000

EID ENAME SALARY

200 manish 100000

201 ramesh 200000

202 ganesh 4000


[Type text] Page 11

203 suresh 85000

204 mahesh 80000

Insert into certified2 values(200,53);









Select * from certified;

EID AID

---- ----------

200 53

200 50

200 51

200 52

202 54

201 51

202 52

202 55

201 50

EID AID

200 53

200 50

200 51

200 52

202 54

201 51

202 52

202 55

201 50

Insert into flights2 values(11,'bengaluru','frankfurt',2000,9,11,60000);

Insert into flights2 values(12,'bengaluru','frankfurt',2000,8,10,50000);


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Insert into flights2 values(33,'bengaluru','newdelhi',1000,9,11,5000);



Select * from flights;

NUM FROME TOO DISTANCE DEPARTS ARRIVES PRICE

---- ---------- ---------- ---------- ---------- ---------- ----------

11 bengaluru frankfurt 2000 9 11 60000


33 bengaluru newdelhi 1000 9 11 5000


55 engaluru newdelhi 800 12 2 6000

NUM FROME TOO DISTANCE DEPARTS ARRIVES PRICE





55 engaluru newdelhi 800 12 2 6000

i) Find the names of aircraft such that all pilots certified to operate them

have salaries more than Rs.80, 000.

SELECT aname, COUNT (eid)

FROM aircraft2 a, certified2 c

WHERE a.aid=c.aid

GROUP BY aname

INTERSECT

SELECT aname, COUNT (e.eid)

FROM aircraft2 a, certified2 c, employees2 e

WHERE a.aid=c.aid and

c.eid=e.eid and

salary>80000

GROUP BY aname;

ANAME COUNT(EID)

---------- ----------

Air india 1

jet 2

kingfisher 2


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ii) For each pilot who is certified for more than three aircrafts, find the eid and the

maximum cruisingrange of the aircraft for which she or he is certified.

SELECT eid, max(cruisingrange)

FROM aircraft2 a, certified2 c

WHERE c.aid=a.aid

GROUP BY eid

HAVING COUNT(c.aid)>3;

EID MAX(CRUISINGRANGE)

------ ------------------

200 30000

iii) Find the names of pilots whose salary is less than the price of the cheapest

route from Bengaluru to Frankfurt.

SELECT distinct ename

FROM employees2 e,certified2 c

WHERE c.eid=e.eid and

salary<(SELECT min(price)

FROM flights2

WHERE frome='bengaluru' and

too='frankfurt');

ENAME

----------

ganesh

iv) For all aircraft with cruisingrange over 1000 Kms, .find the name of the

aircraft and the average salary of all pilots certified for this aircraft.

SELECT aname,avg(salary)

FROM aircraft2 a, employees2 e, certified2 c


a.aid=c.aid and

cruisingrange>1000

GROUP BY aname;


[Type text] Page 14

ANAME AVG (SALARY)

---------- -----------

Air india 100000

IA 52000

boeing3 4000

boeing5 4000

Kingfisher 150000

v) Find the names of pilots certified for some Boeing aircraft.

SELECT distinct ename

FROM aircraft2 a,certified2 c,employees2 e


a.aid=c.aid and

aname like 'boeing%';

ENAME

-------

Ganesh

vi) Find the aids of all aircraft that can be used on routes from Bengaluru to New

Delhi.

SELECT aid

FROM aircraft2

WHERE cruisingrange>=(SELECT min(distance)

FROM flights2

WHERE frome='bengaluru' and too='newdelhi');

AID

---

50

52

53

54

55


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STUDENT ENROLLMENT DATABASE

Create table Student3 (

Regno varchar(10),

Name varchar(10) not null,

Major varchar(10) not null,

Bdate date,

primary key (Regno));

Create table Course3 (

Course_id integer,

Cname varchar(10) not null,

Dept varchar(10) not null,

primary key (Course_id));

Create table Enroll3 (

Regno varchar(10),

Course_id integer,

Sem integer not null,

Marks integer,

primary key (Regno,Course_id),

foreign key (Regno) references Student3(Regno),

foreign key (Course_id) references Course3(Course_id));

Create table Text3(

ISBN integer,

Booktitle varchar(20) not null,

Publisher varchar(20),

Author varchar(15),

primary key (ISBN));

Create table Book_adoption3(

Courseno integer,

Sem integer,

ISBN integer,

primary key (Courseno,Sem),

foreign key (Courseno) references Course3(Course_id),

foreign key (ISBN) references Text3(ISBN));


[Type text] Page 16

Desc student;

Name Null? Type

------------------ ------------------ -------------

REGNO NOT NULL VARCHAR2 (10)

NAME NOT NULL VARCHAR2 (10)

MAJOR NOT NULL VARCHAR2 (10)

BDATE DATE

Desc course;

Name Null? Type

------------------------- -------- ------------

COURSE_ID NOT NULL NUMBER(4)

CNAME VARCHAR2(10)

DEPT VARCHAR2(5)

Desc Enroll;

Name Null? Type

----------------------------------------- -------- ------------

REGNO NOT NULL VARCHAR2(5)

COURSE_ID NOT NULL NUMBER(4)

SEM NUMBER(2)

MARKS NUMBER(3)

Desc Text;

Name Null? Type

----------------------------------------- -------- -------------

ISBN NOT NULL NUMBER(38)

BOOKTITLE NOT NULL VARCHAR2(20)

PUBLISHER VARCHAR2(20)

AUTHOR VARCHAR2(15)


[Type text] Page 17

Desc book_adoption;

Name Null? Type

----------------------------------------- -------- -----------

COURSENO NOT NULL NUMBER(38)

SEM NOT NULL NUMBER(38)

ISBN NUMBER(38)

Insert into Student3 values('1BI02CS010','Karan','CSE','02-Jan-1984');

Insert into Student3 values('1BI02EE015','Jack','EEE','15-Apr-1983');

Insert into Student3 values('1BI00CS010','Adi','CSE','02-Jan-1982');

Insert into Student3 values('1BI01EC089','Rahul','ECE','01-Dec-1983');

Insert into Student3 values('1BI01ME075','Sachin','MECH','18-Jul-1983');

Select * from Student3;

REGNO NAME MAJOR BDATE

------------------ ---------- ---------- -------------

1BI01ME075 Sachin MECH 18-JUL-83

1BI02CS010 Karan CSE 02-JAN-84

1BI02EE015 Jack EEE 15-APR-83

1BI00CS010 Adi CSE 02-JAN-82

1BI01EC089 Rahul ECE 01-DEC-83

REGNO NAME MAJOR BDATE

1BI01ME075 Sachin MECH 18-JUL-83

1BI02CS010 Karan CSE 02-JAN-84

1BI02EE015 Jack EEE 15-APR-83

1BI00CS010 Adi CSE 02-JAN-82

1BI01EC089 Rahul ECE 01-DEC-83

COURSE

Insert into course3 values(11,'DSC','CSE');

Insert into course3 values(22,'ADA','CSE');

Insert into course3 values(33,'CN','EC');

Insert into course3 values(44,'TD','MECH');

Insert into course3 values(55,'MP','EC');


[Type text] Page 18

Select * from course3;

COURSENO CNAME DEPT

------------------- ---------- ------------

11 DSC CSE

22 ADA CSE

33 CN EC

44 TD MECH

55 MP EC

COURSENO CNAME DEPT

11 DSC CSE

22 ADA CSE

33 CN EC

44 TD MECH

55 MP EC

ENROLL

Insert into enroll3 values('1BI02CS010',22,5,72);

Insert into enroll3 values('1BI00CS010',11,3,90);

Insert into enroll3 values('1BI01EC089',33,6,52);

Insert into enroll3 values('1BI01ME075',44,4,85);

Insert into enroll3 values('1BI02EE015',22,5,75);

Select * from enroll3;

REGNO COURSENO SEM MARKS

------------- ----------------- ---------- ------------

1BI02CS010 22 5 72

1BI00CS010 11 3 90

1BI01EC089 33 6 52

1BI01ME075 44 4 85

1BI02EE015 22 5 75

REGNO COURSENO SEM MARKS

1BI02CS010 22 5 72

1BI00CS010 11 3 90

1BI01EC089 33 6 52


[Type text] Page 19

1BI01ME075 44 4 85

1BI02EE015 22 5 75

TEXT

Insert into text3 values(7722,'VB6','Dreamtech','Holzner');

Insert into text3 values(1144,'DS with C','Sapna','Nandagopalan');

Insert into text3 values(4400,'C Programming','TMH','Balaguruswamy');

Insert into text3 values(5566,'Computer Nw','PHI','Tennenbaum');

Insert into text3 values(3388,'MP','PHI','Brey');

Select * from text3;

ISBN BOOKTITLE PUBLISHER AUTHOR

---------- -------------------- -------------------- ------------------

7722 VB6 Dreamtech Holzner

1144 DS with C Sapna Nandagopalan

4400 C Programming TMH Balaguruswamy

5566 Computer Nw PHI Tennenbaum

3388 MP PHI Brey






3388 MP PHI Brey

BOOK_ADOPTION

Insert into book_adoption3 values(11,3,7722);








[Type text] Page 20

Select * from book_adoption3;

COURSENO SEM ISBN

-------------------- ---------- ----------

11 3 7722

22 4 7722

11 5 4400

11 8 5566

55 4 3388

44 4 5566

44 7 3388

COURSENO SEM ISBN

11 3 7722

22 4 7722

11 5 4400

11 8 5566

55 4 3388

44 4 5566

44 7 3388

iii. Demonstrate how you add a new text book to the database and make this book

be adopted by some department.

Insert into text3 values(1234,'Elec.Circuits','Sapna','Giridhar');

1 row created.


1 row created.

Select * from text3;


---------- -------------------- ------------------- ---------------



[Type text] Page 21




3388 MP PHI Brey

1234 Elec.Circuits Sapna Giridhar

6 rows selected.






3388 MP PHI Brey

1234 Elec.Circuits Sapna Giridhar

Select * from book_adoption3;

COURSENO SEM ISBN

-------------------- ---------- ----------

11 3 7722

22 4 7722

11 5 4400

11 8 5566

55 4 3388

44 4 5566

44 7 3388

55 3 1234

8 rows selected.

COURSENO SEM ISBN

11 3 7722

22 4 7722

11 5 4400

11 8 5566

55 4 3388


[Type text] Page 22

44 4 5566

44 7 3388

55 3 1234

iv. Produce a list of text books in alphabetical order for courses offered by CS

department that use more than two books.

SELECT c.Course_id, b.ISBN, Booktitle

FROM Course3 c,Book_adoption3 b,Text3 t

WHERE c.Course_id=b.Courseno and

t.ISBN=b.ISBN and

c.Dept='CSE' and

c.Course_id in (select courseno

from book_adoption

GROUP BY Courseno

Having count(*)>=2)

Order by c.Course_id, Booktitle;

COURSENO ISBN BOOKTITLE

-------------------- ---------- -------------------

11 4400 C Programming

11 5566 Computer Nw

11 7722 VB6

V. List any department that has all its adopted books published by a specific

publisher

Select distinct C.Dept

from Course C, Book_adoption A,Text T

where C.Courseno=A.Courseno and

A.ISBN=T.ISBN and

not exists (( select Y.ISBN

from Course X,Book_Adoption Y

where X.Courseno=Y.Courseno

and X.Dept=C.Dept)

minus

(select ISBN


[Type text] Page 23

from Text

where publisher='PHI'));

DEPT

----------

MECH


[Type text] Page 24

BOOK DEALER DATABASE

Create table Author4(

Authorid integer,

Aname varchar(15),

Acity varchar(15),

Acountry varchar(15),

primary key (Authorid));

Create table Publisher4(

Publisherid integer,

Pname varchar(15),

Pcity varchar(15),

Pcountry varchar(15),

primary key (Publisherid));

Create table Category4(

Categoryid integer,

Description varchar(20),

primary key (Categoryid));

Create table Catalog4(

Bookid integer,

Title varchar(20),

Authorid integer,

Publisherid integer,

Categoryid integer,

Year integer,

Price integer,

primary key (Bookid),

foreign key (Authorid) references Author4(Authorid),

foreign key (Publisherid) references Publisher4(Publisherid),

foreign key (Categoryid) references Category4(Categoryid));

Create table Order_details4(

Orderno integer,

Bookid integer,

Quantity integer,

primary key (Orderno,Bookid),

foreign key (Bookid) references Catalog4(Bookid));


[Type text] Page 25

desc author;

Name Null? Type

------------------ -------- ---------------

AUTHORID NOT NULL NUMBER(38)

ANAME VARCHAR2(15)

ACITY VARCHAR2(15)

ACOUNTRY VARCHAR2(15)

Desc publisher;

Name Null? Type

----------------------------------------- -------- --------------

PUBLISHERID NOT NULL NUMBER(38)

PNAME VARCHAR2(15)

PCITY VARCHAR2(15)

PCOUNTRY VARCHAR2(15)

Desc catalog;

Name Null? Type

----------------------------------------- -------- -------------

BOOK_ID NOT NULL NUMBER(38)

TITLE VARCHAR2(20)

AUTHOR_ID NUMBER(38)

PUBLISHER_ID NUMBER(38)

CATEGORY_ID NUMBER(38)

YEAR NUMBER(38)

PRICE NUMBER(38)

Desc category;

Name Null? Type

----------------------------------------- -------- -------------

CATEGORY_ID NOT NULL NUMBER(38)

DESCRIPTION VARCHAR2(20)

Desc order_details;


[Type text] Page 26

Name Null? Type

----------------------- -------- ------------

ORDER_NO NOT NULL NUMBER(38)

BOOK_ID NOT NULL NUMBER(38)

QUANTITY NUMBER(38)

AUTHOR

Insert into Author4 values(1000,'Nandagopalan','Bangalore','India');

Insert into Author4 values(2000,'Tony','Haywood','USA');

Insert into Author4 values(3000,'Holzner','New York','USA');

Insert into Author4 values(4000,'Tennenbaum','London','UK');

Insert into Author4 values(5000,'Balaguruswamy','Chennai','India');

select * from Author4;

AUTHORID ANAME ACITY ACITY

------------------- -------------- ------------- --------------------

1000 Nandagopalan Bangalore India

2000 Tony Haywood USA

3000 Holzner New York USA

4000 Tennenbaum London UK

5000 Balaguruswamy Chennai India

AUTHORID ANAME ACITY ACITY

1000 Nandagopalan Bangalore India

2000 Tony Haywood USA

3000 Holzner New York USA

4000 Tennenbaum London UK

5000 Balaguruswamy Chennai India

PUBLISHER

Insert into publisher4 values(11,'Wiely','NewDelhi','India');

Insert into publisher4 values(22,'PHI','California','USA');

Insert into publisher4 values(33,'Sapna','Bangalore','India');

Insert into publisher4 values(44,'TMH','NewYork','USA');

Insert into publisher4 values(55,'Wrox','Texas','USA');


[Type text] Page 27

Select * from publisher4;

PUBLISHERID PNAME PCITY PCOUNTRY

--------------------- --------------- --------------- -------------------

11 Wiely NewDelhi India

22 PHI California USA

33 Sapna Bangalore India

44 TMH NewYork USA

55 Wrox Texas USA

PUBLISHERID PNAME PCITY PCOUNTRY

11 Wiely NewDelhi India

22 PHI California USA

33 Sapna Bangalore India

44 TMH NewYork USA

55 Wrox Texas USA

CATEGORY

Insert into category4 values(1,'OS');

Insert into category4 values(2,'Languages');

Insert into category4 values(3,'Hardware');

Insert into category4 values(4,'Algorithms');

Insert into category4 values(5,'Internet');

Select * from category4;

CATEGORYID DESCRIPTION

--------------------- ---------------------

1 OS

2 Languages

3 Hardware

4 Algorithms

5 Internet

CATEGORYID DESCRIPTION

1 OS

2 Languages


[Type text] Page 28

3 Hardware

4 Algorithms

5 Internet

CATALOG

Insert into catalog4 values(222,'DSC',1000,33,2,2000,32);

Insert into catalog4 values(444,'Networks',4000,44,4,2002,365);

Insert into catalog4 values(555,'VB6',2000,11,2,2000,300);

Insert into catalog4 values(333,'Frontpage2002',4000,44,5,2003,500);

Insert into catalog4 values(111,'ADA',1000,33,4,2001,35);

Select * from catalog4;

BOOKID TITLE AUTHORID PUBLISHERID CATEGORYID

YEAR PRICE

222 DSC 1000 33 2

2000 32

444 Networks 4000 44 4

2002 365

555 VB6 2000 11 2

2000 300

333 Frontpage2002 4000 44 5

2003 500

111 ADA 1000 33 4

2001 35

BOOKID TITLE AUTHORID PUBLISHERID CATEGORYID YEAR PRI

CE

222 DSC 1000 33 2 2000 32

444 Networks 4000 44 4 2002 365

555 VB6 2000 11 2 2000 300

333 Frontpage2002 4000 44 5 2003 500


[Type text] Page 29

111 ADA 1000 33 4 2001 35

ORDER_DETAILS

Insert into order_details4 values(112,222,100);





Select * from order_details4;

ORDERNO BOOKID QUANTITY

------------------ -------------- ---------------

112 222 100

113 333 20

114 555 50

115 222 500

116 444 8

ORDERNO BOOKID QUANTITY112

112 222 100

113 333 20

114 555 50

115 222 500

116 444 8

iii) Give the details of the authors who have 2 or more books in the catalog and the

price of the books is greater than the average price of the books in the catalog and

the year of publication is after 2000.

select C.Authorid,a.aname,a.acountry,a.acity

from Catalog4 C,Author4 a

where A.Authorid=C.Authorid

and C.Year>2000

and C.Price>(Select Avg(Price)

from Catalog4)


[Type text] Page 30

group by C.Authorid,a.aname,a.acountry,a.acity

having count(C.Authorid)>=2;

AUTHOR_ID NAME COUNTRY CITY

---------- -------------------- -------------------- --------------------

4000 Tennenbaum UK London

(iv) Find the author of the book which has maximum sales.

Select a.aname,c.authorid,c.bookid

from author4 a,catalog4 c,order_details4 o

where c.bookid=o.bookid

and c.authorid=a.authorid

and o.quantity=(select max(quantity)

from order_details4);

NAME AUTHOR_ID BOOK_ID

-------------------- ---------- ----------

Nandagopal 1000 222

(v) Demonstrate how you increase the price of books published by a specific

publisher by 10%.

UPDATE catalog4

Set price=price+price*0.1

where publisherid in ( SELECT publisherid

from publisher4

where pname='Sapna');

2 rows updated.

or

update catalog

set price=price*1.10

where publisherid=33;

2 rows updated.


[Type text] Page 31

5.BANKING ENTERPRISE

Create table branch5(

branchname varchar(20),

branchcity varchar(20),

assets real,

primary key(branchname));

Create table account5(

accno int primary key,

branchname varchar(20) references branch5(branchname),

balance real);

Create table customer5(

customername varchar(20),

customerstreet varchar(20),

customercity varchar(20),

primary key(customername));

Create table depositor5(


accno int,

primary key(customername,accno),

foreign key(customername) references customer5(customername),

foreign key(accno) references account5(accno)on delete cascade);

Create table loan5(

loanno int,

branchname varchar(20),

amount real,

primary key(loanno),

foreign key(branchname) references branch5(branchname));

Create table borrower5(


loanno int,


[Type text] Page 32

primary key(customername,loanno),

foreign key(customername) references customer5(customername),

foreign key(loanno) references loan5(loanno));

DESC branch;

Name Null? Type

---------------------------------- ----------------- -----------------------------

BRANCHNAME NOT NULL VARCHAR2(15)

BRANCHCITY VARCHAR2(15)

ASSETS NUMBER(63)

DESC accnt;

Name Null? Type

----------------------------------- -------------- ---------------

ACCOUNTNO NOT NULL NUMBER(38)

BRANCHNAME VARCHAR2(15)

BALANCE NUMBER(63)

DESC custmer;

Name Null? Type

---------------------------------- ---------------- -----------------

CUSTOMERNAME NOT NULL VARCHAR2(15)

CUSTOMERSTREET VARCHAR2(15)

CITY VARCHAR2(15)

DESC depositor;

Name Null? Type

-------------------------------- ----------------- ----------------------------


ACCOUNTNO NOT NULL NUMBER(38)

desc loan;

Name Null? Type

---------------------------- ---------------- ----------------------------

LOANNO NOT NULL NUMBER(38)


[Type text] Page 33

BRANCHNAME VARCHAR2(15)

AMOUNT NUMBER(63)

DESC borrower;

Name Null? Type

----------------------------- ---------- ----------------------------------


LOANNO NOT NULL NUMBER(38)

Insert into branch5 values('Jayanagar','Bangalore','15000000');

Insert into branch5 values('Basavanagudi','Bangalore','25000000');

Insert into branch5 values('Noida','NewDelhi','50000000');

Insert into branch5 values('Marinedrive','Mumbai','40000000');

Insert into branch5 values('GreenPark','Newdelhi','30000000');

Select * from branch5;

BRANCHNAME BRANCHCITY ASSETS

--------------- --------------- ----------

Jayanagar Bangalore 15000000

Basavanagudi Bangalore 25000000

Noida NewDelhi 50000000

Marinedrive Mumbai 40000000

GreenPark Newdelhi 30000000

BRANCHNAME BRANCHCITY ASSETS

ayanagar Bangalore 15000000

Basavanagudi Bangalore 25000000

Noida NewDelhi 50000000

Marinedrive Mumbai 40000000

GreenPark NewDelhi 30000000

Insert into account5 values('123','Jayanagar','25000');

Insert into account5 values('156','Jayanagar','30000');

Insert into account5 values('456','Basavanagudi','15000');

Insert into account5 values('789','Noida','25000');

Insert into account5 values('478','Marinedrive','48000');

Insert into account5 values('778','GreenPark','60000');


[Type text] Page 34

Insert into account5 values('189','Basavanagudi','48888');

Select * from account5;

ACCOUNTNO BRANCHNAME BALANCE

--------- --------------- ----------

123 Jayanagar 25000

156 Jayanagar 30000

456 Basavanagudi 15000

789 Noida 25000

478 Marinedrive 48000

778 GreenPark 60000


ACCOUNTNO BRANCHNAME BALANCE

123 Jayanagar 25000

156 Jayanagar 30000


789 Noida 25000


778 GreenPark 60000


Insert into customer5 values('Ramu','Jayanagar','Bangalore');

Insert into customer5 values('Kumar','Basavanagudi','Bangalore');

Insert into customer5 values('John','Noida','Newdelhi');

Insert into customer5 values('Mike','Marinedrive','Mumbai');

Insert into customer5 values('Sachin','GreenPark','NewDelhi');

Select * from customer5;

CUSTOMERNAME CUSTOMERSTREET CITY

--------------- --------------- -------------

Ramu Jayanagar Bangalore

Kumar Basavanagudi Bangalore

John Noida Newdelhi

Mike Marinedrive Mumbai

Sachin GreenPark NewDelhi


[Type text] Page 35

CUSTOMERNAME CUSTOMERSTREET CITY

Ramu Jayanagar Bangalore

Kumar Basavanagudi Bangalore

John Noida Newdelhi

Mike Marinedrive Mumbai

Sachin GreenPark Newdelhi

Insert into depositor5 values('Ramu',123);



Insert into depositor5 values('Kumar',456);

Insert into depositor5 values('John',789);

Insert into depositor5 values('Mike',478);

Insert into depositor5 values('Sachin',778);

Select * from depositor5;

CUSTOMERNAME ACCOUNTNO

--------------- ----------

Ramu 123

Ramu 156

Ramu 189

Kumar 456

John 789

Mike 478

Sachin 778

CUSTOMERNAME ACCOUNTNO

Ramu 123

Ramu 156

Ramu 189

Kumar 456

John 789

Mike 478

Sachin 778

Insert into loan5 values('1111','Jayanagar','250000');


[Type text] Page 36

Insert into loan5 values('2222','Basavanagudi','350000');

Insert into loan5 values('3333','Noida','150000');

Insert into loan5 values('4444','Marinedrive','1500000');

Insert into loan5 values('5555','GreenPark','7500000');

Select * from loan5;

LOANNO BRANCHNAME AMOUNT

------- --------------- ----------

1111 Jayanagar 250000


3333 Noida 150000


5555 GreenPark 7500000

LOANNO BRANCHNAME AMOUNT

1111 Jayanagar 250000


3333 Noida 150000


5555 GreenPark 7500000

BORROWER

Insert into borrower5 values('Ramu',1111);

Insert into borrower5 values('Kumar',2222);

Insert into borrower5 values('John',3333);

Insert into borrower5 values('Mike',4444);

Insert into borrower5 values('Sachin',5555);

Select * from borrower5;

CUSTOMERNAME LOANNO

--------------- ----------

Ramu 1111

Kumar 2222

John 3333

Mike 4444

Sachin 5555


[Type text] Page 37

CUSTOMERNAME LOANNO

Ramu 1111

Kumar 2222

John 3333

Mike 4444

Sachin 5555

(iii) Find all the customers who have at least two accounts at the main branch.

Select d.customername

from depositor5 d , branch5 b, account5 a

where b.branchname='Jayanagar'

and b.branchname = a.branchname

and a.accno = d.accno

group by d.customername

having count(*)>1;

CUSTOMERNAME

---------------

Ramu

(iv)Find all the customers who have an account at all the branches located in a

specific city.

Select d.customername

from account5 a,branch5 b,depositor5 d

where a.accno=d.accno

and b.branchname = a.branchname

and b.branchcity='Bangalore'

group by d.customername

having count(unique(b.branchname)) = (select count(b.branchname)

from branch5 b

where b.branchcity='Bangalore');

CUSTOMERNAME

---------------

Ramu


[Type text] Page 38

Delete from account5

where branchname IN (select b.branchname

from branch5 b

where b.branchcity = 'Bangalore');

4 rows deleted.

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